Exercício Dinâmica - Engenharia Mecânica (1895)

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675 
PROBLEM 13.111* (Continued) 
 
Thus, additional kinetic energy at A is 
 
6
2
110
1 (254.46 10 )
( ) ft lb
2 2A
m
m v E
×Δ = Δ = ⋅ (1) 
Conservation of energy between A and B: 
 2 2
circ
1
[( ) ( ) ]
2A A A A
A
GMm
T m v v V
r
−= + Δ = 
 21
2B B B
A
GMm
T mv V
r
−= = 
 A A B BT V T V+ = + 
 
6 15 15
3 2 2
6 6
1 254.46 10 14.077 10 1 14.077 10
(25.24 10 )
2 2 222.097 10 21.120 10
B
m m m
m mv
× × ×× + − = −
× ×
 
 
2 6 6 6 6
2 3
637.06 10 254.46 10 1274.1 10 1333 10
950.4 10
B
B
v
v
= × + × − × + ×
= ×
 
 330.88 10 ft/sBv = × 
Conservation of angular momentum between A and B: 
 circ( ) sinA A B B Br m v r mv φ= 
 
3
circ
3
( ) (4185) (25.24 10 )
sin 0.8565
( ) (4000) (30.88 10 )
AA
B
B B
vr
r v
φ
  ×= = = 
× 
 
 58.9Bφ = ° 